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Solving an area problem in Multivariable Calculus, Polar mode

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the average area of an inscribed triangle in the unit circle. Assume that each vertex of the triangle is equally likely to be at any point of the unit circle and that the location of one vertex does not affect the likelihood the location of another in any way. (Note that, as seen in Problem set 4, the maximum area is achieved by the equilateral triangle, which has
    side length √3 and area 3√3/4. How does the maximum compare to the average?)
    Hint: in order to reduce the problem to the calculation of a double integral, place one of the vertices of the triangle at(1,0),and use the polar angles θ1 and θ2 of the two other vertices as variables. What is the region of integration?

    Thanks

    http://ocw.mit.edu/NR/rdonlyres/Math...nments/ps7.pdf [Broken]

    (I'm not at MIT, I'm taking the class online, and this problem REALLY irritated me)

    2. Relevant equations

    Area of a parallelogram is the cross product of two vectors, which I said were <rcosø – 1, rsinø> and <rcosß – 1, sinß>. Since we have a unit circle, I make this <cosø – 1, sinø> and <cosß – 1, sinß>; the area of the triangle formed would then be that cross product divided by 2.

    3. The attempt at a solution

    My attempt was to say that dA = r^2 * dødß, or, in the unit circle, dødß

    My cross product was: sinß(cosø–1) + sinø(1–cosß). I took the absolute value of that, then called that the area, integrated it and then divided by the rectangle that the bounds would form, to get the average value of the area:

    1/4π2[∫∫(sinß(1–cosø) + sinø(1–cosß))dødß] with bounds of 0 to π and 0 to π to make sure it is always positive, I changed the bounds and changed sinß(cosø–1) to sinß(1–cosø). Since the bounds would usually be 0 to 2π, I multiplied through by 22 = 4 because I would get the same result if I then made everything negative and added the same integral from π to 2π (to make sure I get absolute value)
    I integrated it down to the ∫ (πsinß + 2 – 2cosß) dß, and then evaluated that to get 4π. I divided that by 2, and then by π2 to get the average triangle area. My answer: ::::::::2/π::::::::

    Am I on the right track or totally wrong???
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 14, 2017 #2
    I'm working through the same online course now, and just finished this problem.

    I arrived at the same cross product, then simplified it further by the angle addition formula to sin(∅-β)+sin(β)-sin(∅).
    From here, to get the average area, you must integrate the function with a double integral, but you have to make sure your integral doesn't cancel itself out. If your bounds for both variables are the same, your integral will evaluate to 0 (plugging in the integral you gave to wolframalpha it confirms this. You morph the integral to fix this, but these changes are artificial). Logically, this makes sense. With double integrals, you are integrating the inner integral while holding the outer variable constant, then increasing the outer variable and repeating the process. So if you hold the one angle at, say 0, and then let the other angle vary from 0 to pi, the area values will be positive, because of how cross-product works. But later in the integral you hold the angle at pi, and all the values are negative, nullifying any calculation you have done.
    To solve this, I set the inner limits to be from β to 2pi, and the outer limits to be from 0 to 2pi. This way the same angle variable always describes the vertex in front. I had a few qualms about doing this because it said to not let the position of one vertex effect the position of another, but this still satisfies that. This hits all the triangles while making sure you don't hit any backwards.
    Evaluating this integral without the integrate, I got 4pi^2. Evaluating the integral then dividing it by 4pi^2, my answer was 3/2pi, or about .47746...

    Obviously I can't say if that logic is completely correct because there are no solutions, but I did find this answer on yahoo answers, and it agrees with my result:
    https://answers.yahoo.com/question/index?qid=20090617080955AALXSAo
     
  4. Jul 17, 2017 #3

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Hi @milksushi . Looking at the date (2009) on the post, I don't expect the person to still be interested in this. Actually, there are several old questions in this practice problem section, which remain unanswered.

    I think many of these Practice Problems should have been posted in the Homework section, instead.
     
  5. Jul 18, 2017 #4
    They are all old homework threads that never got a reply :)
     
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